Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Shear force & bending moment

Shear and moment: V = ΣF on one side, M = ΣF·d on one side
Bending stress: σ = M·y/I
Rectangle: I = bh³/12
Section modulus: W = I/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid: ȳ = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Beam deflection

Point load at midspan: δ = PL³/(48EI)
Uniform load, simply supported: δ = 5wL⁴/(384EI)
Cantilever, end load: δ = PL³/(3EI)
Cantilever, uniform load: δ = wL⁴/(8EI)
Superposition: δ = δ₁ + δ₂

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 50 mm, E = 110 GPa). Both are 300 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.

P = 250 kNrod: Ø20 mm, E = 200 GPatube: Ø120/Ø50 mm, E = 110 GPaL = 300 mm

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø120/Ø70 mm, E = 120 GPa, σ_allow = 80 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid aluminium shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 250 N·md = 30 mmL = 2.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 75 kW at 300 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 75 kWN = 300 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mw = 5 kN/m from 0 to 6 mx = 3 m

M_x = kN·m

6.Shear force & bending momentHard

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

15 kN @ 0 m25 kN @ 6 m50 kN @ 3 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 290 mm and inside diameter d = 258 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 290 mmd = 258 mm

r = mm

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 30 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 240 mmt_f = 10 mmt_w = 12 mmh_w = 220 mmM = 30 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −10 MPa, σᵧ = −40 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −10 MPaσ_y = −40 MPaτ_xy = 30 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 800 mm and wall thickness 10 mm holds gas at a gauge pressure of 0.5 MPa. It is made from plate welded along a helical seam at 35° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 800 mmt = 10 mmp = 0.5 MPaα = 35°

σ_n = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δ_P under the load.

P = 19 kNa = 1 mL = 3 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 125 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/300. Find the smallest depth h_req of the section.

w = 2.5 kN/mL = 4.5 mE = 11 GPa (timber)b = 125 mm

h_req = mm

13.Column bucklingMedium

A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 1.5 mE = 70 GPa (aluminium)b × h = 25 × 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 60 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 60 kNn = 2L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.06875 mm

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·9346 = 1.091 × 10⁹ N

Shortening

δ = 250000·300/1.091 × 10⁹ = 0.06875 mm

Answer

δ = 0.06875 mm

2.Axial stress & strain

Answer: P_max = 638.8 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 120·10³·7461 = 9.582 × 10⁸ N

Governing strain

ε = min(0.001, 0.0006667) = 0.0006667 (copper)

Largest load

P_max = 0.0006667·9.582 × 10⁸ = 638791 N = 638.8 kN

Answer

P_max = 638.8 kN

3.Torsion of shafts

Answer: φ = 17.32 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Polar second moment of area

J = π·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·2500/(26000·79522) = 0.3023 rad

φ = 0.3023·180/π = 17.32°

Answer

φ = 17.32°

4.Torsion of shafts

Answer: d = 67.24 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 75000/31.42 = 2387 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·2.387 × 10⁶/(π·40)) = 67.24 mm

Answer

d = 67.24 mm

In practice you would round up to the next standard size.

5.Shear force & bending moment

Answer: M_x = 22.5 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·6 = 30 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 5·6·3

R_B = 90/6 = 15 kN

ΣF_y = 0: R_A = 30 − 15 = 15 kN

Cut at x = 3 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 15·3 − 5·3·1.5 = 22.5 kN·m

Answer

M_x = 22.5 kN·m

6.Shear force & bending moment

Answer: E (x = 4 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 20 kN, R₂ = 70 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = −15, M_C = −10, M_D = −5, M_E = −50 kN·m, M_F = −25, M_G = 0

Answer

Mark E: M_E = −50 kN·m

7.Centroid & second moment of area

Answer: r = 97.04 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 13773 mm²

I = π(D⁴ − d⁴)/64 = 1.297 × 10⁸ mm⁴

Radius of gyration

r = √(I/A) = √(1.297 × 10⁸/13773) = 97.04 mm

Answer

r = 97.04 mm

8.Centroid & second moment of area

Answer: σ_top = 71.71 MPa

Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 240·10 = 2400 mm², y_f = 220 + 10/2 = 225 mm

web: A_w = 12·220 = 2640 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2400·225 + 2640·110)/5040 = 164.8 mm

Parallel-axis theorem

flange: 240·10³/12 + 2400·(225 − 164.8)² = 8.729 × 10⁶ mm⁴

web: 12·220³/12 + 2640·(110 − 164.8)² = 1.857 × 10⁷ mm⁴

I_x = 8.729 × 10⁶ + 1.857 × 10⁷ = 2.729 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 230 − 164.8 = 65.24 mm

σ = M·y/I = 3 × 10⁷·65.24/2.729 × 10⁷ = 71.71 MPa

Answer

σ_top = 71.71 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 33.54 MPa

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

Radius of Mohr’s circle

τ_max = R = √(((−10 − −40)/2)² + 30²) = 33.54 MPa

Answer

τ_max = 33.54 MPa

10.Mohr’s circle & stress transformation

Answer: σ_n = 16.71 MPa

Wall stresses

r = 800/2 = 400 mm

σ_h = pr/t = 0.5·400/10 = 20 MPa

σ_a = pr/(2t) = 10 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 10·sin²35° + 20·cos²35° = 16.71 MPa

Answer

σ_n = 16.71 MPa

11.Beam deflection

Answer: δ_P = 6.397 mm

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

P = 19000 N, a = 1000 mm, b = 2000 mm, L = 3000 mm

EI = 1.32 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 19000·1000²·2000²/(3·1.32 × 10¹²·3000) = 6.397 mm

Answer

δ_P = 6.397 mm

12.Beam deflection

Answer: h_req = 198 mm

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

w = 2.5 N/mm, L = 4500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/300 = 4500/300 = 15 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2.5·4500⁴/(384·11000·15) = 8.09 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·8.09 × 10⁷/125)^1/3 = 198 mm

Answer

h_req = 198 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 15.99 kN

Euler buckling: a slender column becomes unstable at P_cr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 40·25³/12 = 52083 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·52083/1500² = 15992 N = 15.99 kN

Answer

P_cr = 15.99 kN

14.Column buckling

Answer: d = 47.18 mm

Required critical load

P_cr = n·P = 2·60000 = 120000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·120000·2000²/(π³·200000))^(1/4) = 47.18 mm

Answer

d = 47.18 mm

Check: the slenderness KL/r is 169.6, so the column is slender and Euler applies.

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